Introduction to Real Analysis. Lee Larson University of Louisville July 23, 2018

Transcription

1 Introduction to Rel Anlysis Lee Lrson University of Louisville

2

3 About This Document I often tech the MATH : Introduction to Rel Anlysis course t the University of Louisville. The course is intended for mix of mostly upper-level mthemtics mjors with smttering of other students from mthemtics, physics nd engineering. These re notes I ve compiled over the yers. They cover the bsic ides of nlysis on the rel line. Prerequisites re good clculus course, including stndrd differentition nd integrtion methods for rel functions, nd course in which the students must red nd write proofs. Some fmilirity with bsic set theory nd stndrd proof methods such s induction nd contrdiction is needed. The most importnt thing is some mthemticl sophistiction beyond the bsic lgorithm- nd computtion-bsed courses. Feel free to use these notes for ny purpose, s long s you give me blme or credit. In return, I only sk you to tell me bout mistkes. Any suggestions for improvements nd dditions re very much pprecited. I cn be contcted using the emil ddress on the Web pge referenced below. The notes re updted nd corrected quite often. The dte of the version you re reding is t the bottom-left of most pges. The ltest version is vilble for downlod t the Web ddress mth.louisville.edu/ lee/ir. There re mny exercises t the ends of the chpters. There is no generl collection of solutions. Some erly versions of the notes leked out onto the Internet nd they re being offered by few of the usul downlod sites. The erly versions were ment for me nd my clsses only, nd contin mny typos nd few gsp! outright mistkes. Plese help me expunge these escpees from the Internet by pointing those who offer the older files to the ltest version. i

4 Contents About This Document i Chpter 1. Bsic Ides Sets Algebr of Sets Indexed Sets Functions nd Reltions Crdinlity Exercises 1-14 Chpter 2. The Rel Numbers The Field Axioms The Order Axiom The Completeness Axiom Comprisons of Q nd R Exercises 2-12 Chpter 3. Sequences Bsic Properties Monotone Sequences Subsequences nd the Bolzno-Weierstrss Theorem Lower nd Upper Limits of Sequence The Nested Intervl Theorem Cuchy Sequences Exercises 3-15 Chpter 4. Series Wht is Series? Positive Series Absolute nd Conditionl Convergence Rerrngements of Series Exercises 4-17 Chpter 5. The Topology of R Open nd Closed Sets Reltive Topologies nd Connectedness Covering Properties nd Compctness on R More Smll Sets 5-9 i

5 5. Exercises 5-15 Chpter 6. Limits of Functions Bsic Definitions Unilterl Limits Continuity Unilterl Continuity Continuous Functions Uniform Continuity Exercises 6-15 Chpter 7. Differentition The Derivtive t Point Differentition Rules Derivtives nd Extreme Points Differentible Functions Applictions of the Men Vlue Theorem Exercises 7-16 Chpter 8. Integrtion Prtitions Riemnn Sums Drboux Integrtion The Integrl The Cuchy Criterion Properties of the Integrl The Fundmentl Theorem of Clculus Chnge of Vribles Integrl Men Vlue Theorems Exercises 8-23 Chpter 9. Sequences of Functions Pointwise Convergence Uniform Convergence Metric Properties of Uniform Convergence Series of Functions Continuity nd Uniform Convergence Integrtion nd Uniform Convergence Differentition nd Uniform Convergence Power Series Exercises 9-25 Chpter 10. Fourier Series Trigonometric Polynomils The Riemnn Lebesgue Lemm The Dirichlet Kernel Dini s Test for Pointwise Convergence 10-7

6 5. Gibbs Phenomenon A Divergent Fourier Series The Fejér Kernel Exercises Appendix. Bibliogrphy 22 Appendix. Index A-2

7

8 CHAPTER 1 Bsic Ides In the end, ll mthemtics cn be boiled down to logic nd set theory. Becuse of this, ny creful presenttion of fundmentl mthemticl ides is inevitbly couched in the lnguge of logic nd sets. This chpter defines enough of tht lnguge to llow the presenttion of bsic rel nlysis. Much of it will be fmilir to you, but look t it nywy to mke sure you understnd the nottion. 1. Sets Set theory is lrge nd complicted subject in its own right. There is no time in this course to touch on ny but the simplest prts of it. Insted, we ll just look t few topics from wht is often clled nive set theory, mny of which should lredy be fmilir to you. We begin with few definitions. A set is collection of objects clled elements. Usully, sets re denoted by the cpitl letters A, B,, Z. A set cn consist of ny type nd number of elements. Even other sets cn be elements of set. The sets delt with here usully hve rel numbers s their elements. If is n element of the set A, we write A. If is not n element of the set A, we write / A. If ll the elements of A re lso elements of B, then A is subset of B. In this cse, we write A B or B A. In prticulr, notice tht whenever A is set, then A A. Two sets A nd B re equl, if they hve the sme elements. In this cse we write A = B. It is esy to see tht A = B iff A B nd B A. Estblishing tht both of these continments re true is the most common wy to show two sets re equl. If A B nd A B, then A is proper subset of B. In cses when this is importnt, it is written A B insted of just A B. There re severl wys to describe set. A set cn be described in words such s P is the set of ll presidents of the United Sttes. This is cumbersome for complicted sets. All the elements of the set could be listed in curly brces s S = {2, 0, }. If the set hs mny elements, this is imprcticl, or impossible. 1-1

9 1-2 CHAPTER 1. BASIC IDEAS More common in mthemtics is set builder nottion. Some exmples re P = {p : p is president of the United sttes} nd = {Wshington, Adms, Jefferson,, Clinton, Bush, Obm, Trump} A = {n : n is prime number} = {2, 3, 5, 7, 11, }. In generl, the set builder nottion defines set in the form {formul for typicl element : objects to plug into the formul}. A more complicted exmple is the set of perfect squres: S = {n 2 : n is n integer} = {0, 1, 4, 9, }. The existence of severl sets will be ssumed. The simplest of these is the empty set, which is the set with no elements. It is denoted s. The nturl numbers is the set N = {1, 2, 3, } consisting of the positive integers. The set Z = {, 2, 1, 0, 1, 2, } is the set of ll integers. ω = {n Z : n 0} = {0, 1, 2, } is the nonnegtive integers. Clerly, A, for ny set A nd N ω Z. Definition 1.1. Given ny set A, the power set of A, written P(A), is the set consisting of ll subsets of A; i.e., P(A) = {B : B A}. For exmple, P({, b}) = {, {}, {b}, {, b}}. Also, for ny set A, it is lwys true tht P(A) nd A P(A). If A, it is rrely true tht P(A), but it is lwys true tht {} P(A). Mke sure you understnd why! An musing exmple is P( ) = { }. (Don t confuse with { }! The former is empty nd the ltter hs one element.) Now, consider P( ) = { } P(P( )) = {, { }} P(P(P( ))) = {, { }, {{ }}, {, { }}} After continuing this n times, for some n N, the resulting set, P(P( P( ) )), is very lrge. In fct, since set with k elements hs 2 k elements in its power set, there re = 65, 536 elements fter only five itertions of the exmple. Beyond this, the numbers re too lrge to print. Number sequences such s this one re sometimes clled tetrtions.

10 2. ALGEBRA OF SETS 1-3 A B A B A B A B A \ B A B A B A B Figure 1.1. These re Venn digrms showing the four stndrd binry opertions on sets. In this figure, the set which results from the opertion is shded. 2. Algebr of Sets Let A nd B be sets. There re four common binry opertions used on sets. 1 The union of A nd B is the set contining ll the elements in either A or B: A B = {x : x A x B}. The intersection of A nd B is the set contining the elements contined in both A nd B: A B = {x : x A x B}. The difference of A nd B is the set of elements in A nd not in B: A \ B = {x : x A x / B}. The symmetric difference of A nd B is the set of elements in one of the sets, but not the other: A B = (A B) \ (A B). Another common set opertion is complementtion. The complement of set A is usully thought of s the set consisting of ll elements which re not in A. But, little thinking will convince you this is not meningful definition becuse the collection of elements not in A is not precisely 1 In the following, some logicl nottion is used. The symbol is the logicl nonexclusive or. The symbol is the logicl nd. Their truth tbles re s follows: T F T T F F F F T F T T T F T F

11 1-4 CHAPTER 1. BASIC IDEAS understood collection. To mke sense of the complement of set, there must be well-defined universl set U which contins ll the sets in question. Then the complement of set A U is A c = U \ A. It is usully the cse tht the universl set U is evident from the context in which it is used. With these opertions, n extensive lgebr for the mnipultion of sets cn be developed. It s usully done hnd in hnd with forml logic becuse the two subjects shre much in common. These topics re studied s prt of Boolen lgebr. 2 Severl exmples of set lgebr re given in the following theorem nd its corollry. Theorem 1.2. Let A, B nd C be sets. () A \ (B C) = (A \ B) (A \ C) (b) A \ (B C) = (A \ B) (A \ C) Proof. () This is proved s sequence of equivlences. 3 x A \ (B C) x A x / (B C) x A x / B x / C (x A x / B) (x A x / C) x (A \ B) (A \ C) (b) This is lso proved s sequence of equivlences. x A \ (B C) x A x / (B C) x A (x / B x / C) (x A x / B) (x A x / C) x (A \ B) (A \ C) Theorem 1.2 is version of pir of set equtions which re often clled De Morgn s Lws. 4 The more usul sttement of De Morgn s Lws is in Corollry 1.3, which is n obvious consequence of Theorem 1.2 when there is universl set to mke complementtion well-defined. Corollry 1.3 (De Morgn s Lws). Let A nd B be sets. () (A B) c = A c B c (b) (A B) c = A c B c 2 George Boole ( ) 3 The logicl symbol is the sme s if, nd only if. If A nd B re ny two sttements, then A B is the sme s sying A implies B nd B implies A. It is lso common to use iff in this wy. 4 Augustus De Morgn ( )

12 4. FUNCTIONS AND RELATIONS Indexed Sets We often hve occsion to work with lrge collections of sets. For exmple, we could hve sequence of sets A 1, A 2, A 3,, where there is set A n ssocited with ech n N. In generl, let Λ be set nd suppose for ech λ Λ there is set A λ. The set {A λ : λ Λ} is clled collection of sets indexed by Λ. In this cse, Λ is clled the indexing set for the collection. Exmple 1.1. For ech n N, let A n = {k Z : k 2 n}. Then A 1 = A 2 =A 3 = { 1, 0, 1}, A 4 = { 2, 1, 0, 1, 2},, A 61 = { 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7}, is collection of sets indexed by N. Two of the bsic binry opertions cn be extended to work with indexed collections. In prticulr, using the indexed collection from the previous prgrph, we define A λ = {x : x A λ for some λ Λ} nd λ Λ A λ = {x : x A λ for ll λ Λ}. λ Λ De Morgn s Lws cn be generlized to indexed collections. Theorem 1.4. If {B λ : λ Λ} is n indexed collection of sets nd A is set, then A \ λ Λ B λ = λ Λ(A \ B λ ) nd A \ λ Λ B λ = λ Λ(A \ B λ ). Proof. The proof of this theorem is Exercise Functions nd Reltions 4.1. Tuples. When listing the elements of set, the order in which they re listed is unimportnt; e.g., {e, l, v, i, s} = {l, i, v, e, s}. If the order in which n items re listed is importnt, the list is clled n n-tuple. (Strictly speking, n n-tuple is not set.) We denote n n-tuple by enclosing the ordered list in prentheses. For exmple, if x 1, x 2, x 3, x 4 re four items, the 4-tuple (x 1, x 2, x 3, x 4 ) is different from the 4-tuple (x 2, x 1, x 3, x 4 ). Becuse they re used so often, the cses when n = 2 nd n = 3 hve specil nmes: 2-tuples re clled ordered pirs nd 3-tuple is clled n ordered triple.

13 1-6 CHAPTER 1. BASIC IDEAS Definition 1.5. Let A nd B be sets. The set of ll ordered pirs A B = {(, b) : A b B} is clled the Crtesin product of A nd B. 5 nd Exmple 1.2. If A = {, b, c} nd B = {1, 2}, then A B = {(, 1), (, 2), (b, 1), (b, 2), (c, 1), (c, 2)}. B A = {(1, ), (1, b), (1, c), (2, ), (2, b), (2, c)}. Notice tht A B B A becuse of the importnce of order in the ordered pirs. A useful wy to visulize the Crtesin product of two sets is s tble. The Crtesin product A B from Exmple 1.2 is listed s the entries of the following tble. 1 2 (, 1) (, 2) b (b, 1) (b, 2) c (c, 1) (c, 2) Of course, the common Crtesin plne from your nlytic geometry course is nothing more thn generliztion of this ide of listing the elements of Crtesin product s tble. The definition of Crtesin product cn be extended to the cse of more thn two sets. If {A 1, A 2,, A n } re sets, then A 1 A 2 A n = {( 1, 2,, n ) : k A k for 1 k n} is set of n-tuples. This is often written s n A k = A 1 A 2 A n Reltions. Definition 1.6. If A nd B re sets, then ny R A B is reltion from A to B. If (, b) R, we write Rb. In this cse, dom (R) = { : (, b) R} A is the domin of R nd rn (R) = {b : (, b) R} B is the rnge of R. It my hppen tht dom (R) nd rn (R) re proper subsets of A nd B, respectively. 5 René Descrtes,

14 4. FUNCTIONS AND RELATIONS 1-7 In the specil cse when R A A, for some set A, there is some dditionl terminology. R is symmetric, if Rb br. R is reflexive, if R whenever dom (A). R is trnsitive, if Rb brc = Rc. R is n equivlence reltion on A, if it is symmetric, reflexive nd trnsitive. Exmple 1.3. Let R be the reltion on Z Z defined by Rb b. Then R is reflexive nd trnsitive, but not symmetric. Exmple 1.4. Let R be the reltion on Z Z defined by Rb < b. Then R is trnsitive, but neither reflexive nor symmetric. Exmple 1.5. Let R be the reltion on Z Z defined by Rb 2 = b 2. In this cse, R is n equivlence reltion. It is evident tht Rb iff b = or b = Functions. Definition 1.7. A reltion R A B is function if Rb 1 Rb 2 = b 1 = b 2. If f A B is function nd dom (f) = A, then we usully write f : A B nd use the usul nottion f() = b insted of fb. If f : A B is function, the usul intuitive interprettion is to regrd f s rule tht ssocites ech element of A with unique element of B. It s not necessrily the cse tht ech element of B is ssocited with something from A; i.e., B my not be rn (f). It s lso common for more thn one element of A to be ssocited with the sme element of B. Exmple 1.6. Define f : N Z by f(n) = n 2 nd g : Z Z by g(n) = n 2. In this cse rn (f) = {n 2 : n N} nd rn (g) = rn (f) {0}. Notice tht even though f nd g use the sme formul, they re ctully different functions. Definition 1.8. If f : A B nd g : B C, then the composition of g with f is the function g f : A C defined by g f() = g(f()). In Exmple 1.6, g f(n) = g(f(n)) = g(n 2 ) = (n 2 ) 2 = n 4 mkes sense for ll n N, but f g is undefined t n = 0. There re severl importnt types of functions. Definition 1.9. A function f : A B is constnt function, if rn (f) hs single element; i.e., there is b B such tht f() = b for ll A. The function f is surjective (or onto B), if rn (f) = B. In sense, constnt nd surjective functions re the opposite extremes. A constnt function hs the smllest possible rnge nd surjective function hs the lrgest possible rnge. Of course, function f : A B cn be both constnt nd surjective, if B hs only one element.

15 1-8 CHAPTER 1. BASIC IDEAS f b A c f f B g b c g A g B Figure 1.2. These digrms show two functions, f : A B nd g : A B. The function g is injective nd f is not becuse f() = f(c). Definition A function f : A B is injective (or one-to-one), if f() = f(b) implies = b. The terminology one-to-one is very descriptive becuse such function uniquely pirs up the elements of its domin nd rnge. An illustrtion of this definition is in Figure 1.2. In Exmple 1.6, f is injective while g is not. Definition A function f : A B is bijective, if it is both surjective nd injective. A bijective function cn be visulized s uniquely piring up ll the elements of A nd B. Some uthors, fvoring less pretentious lnguge, use the more descriptive terminology one-to-one correspondence insted of bijection. This piring up of the elements from ech set is like counting them nd finding they hve the sme number of elements. Given ny two sets, no mtter how mny elements they hve, the intuitive ide is they hve the sme number of elements if, nd only if, there is bijection between them. The following theorem shows tht this property of counting the number of elements works in fmilir wy. (Its proof is left s n esy exercise.) Theorem If f : A B nd g : B C re bijections, then g f : A C is bijection Inverse Functions.

16 4. FUNCTIONS AND RELATIONS 1-9 f f 1 f 1 A f B Figure 1.3. This is one wy to visulize generl invertible function. First f does something to nd then f 1 undoes it. Definition If f : A B, C A nd D B, then the imge of C is the set f(c) = {f() : C}. The inverse imge of D is the set f 1 (D) = { : f() D}. Definitions 1.11 nd 1.13 work together in the following wy. Suppose f : A B is bijective nd b B. The fct tht f is surjective gurntees tht f 1 ({b}). Since f is injective, f 1 ({b}) contins only one element, sy, where f() = b. In this wy, it is seen tht f 1 is rule tht ssigns ech element of B to exctly one element of A; i.e., f 1 is function with domin B nd rnge A. Definition If f : A B is bijective, the inverse of f is the function f 1 : B A with the property tht f 1 f() = for ll A nd f f 1 (b) = b for ll b B. 6 There is some mbiguity in the mening of f 1 between 1.13 nd The former is n opertion working with subsets of A nd B; the ltter is function working with elements of A nd B. It s usully cler from the context which mening is being used. Exmple 1.7. Let A = N nd B be the even nturl numbers. If f : A B is f(n) = 2n nd g : B A is g(n) = n/2, it is cler f is bijective. Since f g(n) = f(n/2) = 2n/2 = n nd g f(n) = g(2n) = 2n/2 = n, we see g = f 1. (Of course, it is lso true tht f = g 1.) Exmple 1.8. Let f : N Z be defined by { (n 1)/2, n odd, f(n) = n/2, n even It s quite esy to see tht f is bijective nd { f 1 2n + 1, n 0, (n) = 2n, n < 0 6 The nottion f 1 (x) for the inverse is unfortunte becuse it is so esily confused with the multiplictive inverse, (f(x)) 1. For discussion of this, see [8]. The context is usully enough to void confusion.

17 1-10 CHAPTER 1. BASIC IDEAS Given ny set A, it s obvious there is bijection f : A A nd, if g : A B is bijection, then so is g 1 : B A. Combining these observtions with Theorem 1.12, n esy theorem follows. Theorem Let S be collection of sets. The reltion on S defined by A B there is bijection f : A B is n equivlence reltion Schröder-Bernstein Theorem. The following theorem is powerful tool in set theory, nd shows tht seemingly intuitively obvious sttement is sometimes difficult to verify. It will be used in Section 5. Theorem 1.16 (Schröder-Bernstein 7 ). Let A nd B be sets. If there re injective functions f : A B nd g : B A, then there is bijective function h : A B. A A 1 A 2 A 3 A 4 A 5 B 1 B 2 B 3 B 4 B 5 f(a) B Figure 1.4. Here re the first few steps from the construction used in the proof of Theorem Proof. Let B 1 = B \ f(a). If B k B is defined for some k N, let A k = g(b k ) nd B k+1 = f(a k ). This inductively defines A k nd B k for ll k N. Use these sets to define à = k N A k nd h : A B s { g 1 (x), x h(x) = à f(x), x A \ Ã. It must be shown tht h is well-defined, injective nd surjective. To show h is well-defined, let x A. If x A \ Ã, then it is cler h(x) = f(x) is defined. On the other hnd, if x Ã, then x A k for some k. Since x A k = g(b k ), we see h(x) = g 1 (x) is defined. Therefore, h is well-defined. 7 Felix Bernstein ( ), Ernst Schröder ( ) This is often clled the Cntor-Schröder-Bernstein or Cntor-Bernstein Theorem, despite the fct tht it ws pprently first proved by Richrd Dedekind ( ).

18 5. CARDINALITY 1-11 To show h is injective, let x, y A with x y. If both x, y à or x, y A \ Ã, then the ssumptions tht g nd f re injective, respectively, imply h(x) h(y). The remining cse is when x à nd y A \ Ã. Suppose x A k nd h(x) = h(y). If k = 1, then h(x) = g 1 (x) B 1 nd h(y) = f(y) f(a) = B \ B 1. This is clerly incomptible with the ssumption tht h(x) = h(y). Now, suppose k > 1. Then there is n x 1 B 1 such tht This implies x = g f g f f g(x }{{} 1 ). k 1 f s nd k g s h(x) = g 1 (x) = f g f f g(x }{{} 1 ) = f(y) k 1 f s nd k 1 g s so tht y = g f g f f g(x }{{} 1 ) A k 1 Ã. k 2 f s nd k 1 g s This contrdiction shows tht h(x) h(y). We conclude h is injective. To show h is surjective, let y B. If y B k for some k, then h(a k ) = g 1 (A k ) = B k shows y h(a). If y / B k for ny k, y f(a) becuse B 1 = B \ f(a), nd g(y) / Ã, so y = h(x) = f(x) for some x A. This shows h is surjective. The Schröder-Bernstein theorem hs mny consequences, some of which re t first bit unintuitive, such s the following theorem. Corollry There is bijective function h : N N N Proof. If f : N N N is f(n) = (n, 1), then f is clerly injective. On the other hnd, suppose g : N N N is defined by g((, b)) = 2 3 b. The uniqueness of prime fctoriztions gurntees g is injective. An ppliction of Theorem 1.16 yields h. To pprecite the power of the Schröder-Bernstein theorem, try to find n explicit bijection h : N N N. 5. Crdinlity There is wy to use sets nd functions to formlize nd generlize how we count. For exmple, suppose we wnt to count how mny elements re in the set {, b, c}. The nturl wy to do this is to point t ech element in succession nd sy one, two, three. Wht we re doing is defining bijective function between {, b, c} nd the set {1, 2, 3}. This ide cn be generlized. Definition Given n N, the set n = {1, 2,, n} is clled n initil segment of N. The trivil initil segment is 0 =. A set S hs crdinlity n, if there is bijective function f : S n. In this cse, we write crd (S) = n.

19 1-12 CHAPTER 1. BASIC IDEAS The crdinlities defined in Definition 1.18 re clled the finite crdinl numbers. They correspond to the everydy counting numbers we usully use. The ide cn be generlized still further. Definition Let A nd B be two sets. If there is n injective function f : A B, we sy crd (A) crd (B). According to Theorem 1.16, the Schröder-Bernstein Theorem, if crd (A) crd (B) nd crd (B) crd (A), then there is bijective function f : A B. As expected, in this cse we write crd (A) = crd (B). When crd (A) crd (B), but no such bijection exists, we write crd (A) < crd (B). Theorem 1.15 shows tht crd (A) = crd (B) is n equivlence reltion between sets. The ide here, of course, is tht crd (A) = crd (B) mens A nd B hve the sme number of elements nd crd (A) < crd (B) mens A is smller set thn B. This simple intuitive understnding hs some surprising consequences when the sets involved do not hve finite crdinlity. In prticulr, the set A is countbly infinite, if crd (A) = crd (N). In this cse, it is common to write crd (N) = ℵ 0. 8 When crd (A) ℵ 0, then A is sid to be countble set. In other words, the countble sets re those hving finite or countbly infinite crdinlity. Exmple 1.9. Let f : N Z be defined s { n+1 f(n) = 2, when n is odd 1 n. 2, when n is even It s esy to show f is bijection, so crd (N) = crd (Z) = ℵ 0. Theorem Suppose A nd B re countble sets. () A B is countble. (b) A B is countble. Proof. () This is consequence of Theorem (b) This is Exercise An lert reder will hve noticed from previous exmples tht ℵ 0 = crd (Z) = crd (ω) = crd (N) = crd (N N) = crd (N N N) = A logicl question is whether ll sets either hve finite crdinlity, or re countbly infinite. Tht this is not so is seen by letting S = N in the following theorem. Theorem If S is set, crd (S) < crd (P(S)). Proof. Noting tht 0 = crd ( ) < 1 = crd (P( )), the theorem is true when S is empty. nught. 8 The symbol ℵ is the Hebrew letter leph nd ℵ0 is usully pronounced leph

20 5. CARDINALITY 1-13 Suppose S. Since {} P(S) for ll S, it follows tht crd (S) crd (P(S)). Therefore, it suffices to prove there is no surjective function f : S P(S). To see this, ssume there is such function f nd let T = {x S : x / f(x)}. Since f is surjective, there is t S such tht f(t) = T. Either t T or t / T. If t T = f(t), then the definition of T implies t / T, contrdiction. On the other hnd, if t / T = f(t), then the definition of T implies t T, nother contrdiction. These contrdictions led to the conclusion tht no such function f cn exist. A set S is sid to be uncountbly infinite, or just uncountble, if ℵ 0 < crd (S). Theorem 1.21 implies ℵ 0 < crd (P(N)), so P(N) is uncountble. In fct, the sme rgument implies ℵ 0 = crd (N) < crd (P(N)) < crd (P(P(N))) < So, there re n infinite number of distinct infinite crdinlities. In 1874 Georg Cntor [5] proved crd (R) = crd (P(N)) > ℵ 0, where R is the set of rel numbers. (A version of Cntor s theorem ppers in Theorem 2.27 below.) This nturlly led to the question whether there re sets S such tht ℵ 0 < crd (S) < crd (R). Cntor spent mny yers trying to nswer this question nd never succeeded. His ssumption tht no such sets exist cme to be clled the continuum hypothesis. The importnce of the continuum hypothesis ws highlighted by Dvid Hilbert t the 1900 Interntionl Congress of Mthemticins in Pris, when he put it first on his fmous list of the 23 most importnt open problems in mthemtics. Kurt Gödel proved in 1940 tht the continuum hypothesis cnnot be disproved using stndrd set theory, but he did not prove it ws true. In 1963 it ws proved by Pul Cohen tht the continuum hypothesis is ctully unprovble s theorem in stndrd set theory. So, the continuum hypothesis is sttement with the strnge property tht it is neither true nor flse within the frmework of ordinry set theory. This mens tht in the stndrd xiomtic development of set theory, the continuum hypothesis, or creful negtion of it, cn be tken s n dditionl xiom without cusing ny contrdictions. The technicl terminology is tht the continuum hypothesis is independent of the xioms of set theory. The proofs of these theorems re extremely difficult nd entire brod res of mthemtics were invented just to mke their proofs possible. Even tody, there re some deep philosophicl questions swirling round them. A more technicl introduction to mny of these ides is contined in the book by Ciesielski [9]. A nontechnicl nd very redble history of the efforts by mthemticins to understnd the continuum hypothesis is the book by Aczel [1]. A shorter, nontechnicl ccount of Cntor s work is in n rticle by Duben [10].

21 1-14 CHAPTER 1. BASIC IDEAS 6. Exercises 1.1. If set S hs n elements for n ω, then how mny elements re in P(S)? 1.2. Is there set S such tht S P(S)? 1.3. Prove tht for ny sets A nd B, () A = (A B) (A \ B) (b) A B = (A \ B) (B \ A) (A B) nd tht the sets A \ B, B \ A nd A B re pirwise disjoint. (c) A \ B = A B c Prove Theorem For ny sets A, B, C nd D, nd (A B) (C D) = (A C) (B D) (A B) (C D) (A C) (B D). Why does equlity not hold in the second expression? 1.6. Prove Theorem Suppose R is n equivlence reltion on A. For ech x A define C x = {y A : xry}. Prove tht if x, y A, then either C x = C y or C x C y =. (The collection {C x : x A} is the set of equivlence clsses induced by R.) 1.8. If f : A B nd g : B C re bijections, then so is g f : A C Prove or give counter exmple: f : X Y is injective iff whenever A nd B re disjoint subsets of Y, then f 1 (A) f 1 (B) = If f : A B is bijective, then f 1 is unique Prove tht f : X Y is surjective iff for ech subset A X, Y \ f(a) f(x \ A) Suppose tht A k is set for ech positive integer k. () Show tht x n=1 ( k=n A k) iff x A k for infinitely mny sets A k. (b) Show tht x n=1 ( k=n A k) iff x A k for ll but finitely mny of the sets A k. The set n=1 ( k=n A k) from () is often clled the superior limit of the sets A k nd n=1 ( k=n A k) is often clled the inferior limit of the sets A k.

22 6. EXERCISES Given two sets A nd B, it is common to let A( B denote the set of ll functions f : B A. Prove tht for ny set A, crd 2 A) = crd (P(A)). This is why mny uthors use 2 A s their nottion for P(A) Let S be set. Prove the following two sttements re equivlent: () S is infinite; nd, (b) there is proper subset T of S nd bijection f : S T. This sttement is often used s the definition of when set is infinite If S is n infinite set, then there is countbly infinite collection of nonempty pirwise disjoint infinite sets T n, n N such tht S = n N T n Without using the Schröder-Bernstein theorem, find bijection f : [0, 1] (0, 1) If f : [0, ) (0, ) nd g : (0, ) [0, ) re given by f(x) = x + 1 nd g(x) = x, then the proof of the Schrŏder-Bernstein theorem yields wht bijection h : [0, ) (0, )? Find function f : R \ {0} R \ {0} such tht f 1 = 1/f Find bijection f : [0, ) (0, ) If f : A B nd g : B A re functions such tht f g(x) = x for ll x B nd g f(x) = x for ll x A, then f 1 = g If A nd B re sets such tht crd (A) = crd (B) = ℵ 0, then crd (A B) = ℵ Using the nottion from the proof of the Schröder-Bernstein Theorem, let A = [0, ), B = (0, ), f(x) = x + 1 nd g(x) = x. Determine h(x) Using the nottion from the proof of the Schröder-Bernstein Theorem, let A = N, B = Z, f(n) = n nd { 1 3n, n 0 g(n) = 3n 1, n > 0. Clculte h(6) nd h(7) Suppose tht in the sttement of the Schröder-Bernstein theorem A = B = Z nd f(n) = g(n) = 2n. Following the procedure in the proof yields wht function h? If {A n : n N} is collection of countble sets, then n N A n is countble.

23 1-16 CHAPTER 1. BASIC IDEAS If A nd B re sets, the set of ll ( functions f : A B is often denoted by B A. If S is set, prove tht crd 2 S) = crd (P(S)) If ℵ 0 crd (S)), then there is n injective function f : S S tht is not surjective If crd (S) = ℵ 0, then there is sequence of pirwise disjoint sets T n, n N such tht crd (T n ) = ℵ 0 for every n N nd n N T n = S.

24 CHAPTER 2 The Rel Numbers This chpter concerns wht cn be thought of s the rules of the gme: the xioms of the rel numbers. These xioms imply ll the properties of the rel numbers nd, in sense, ny set stisfying them is uniquely determined to be the rel numbers. The xioms re presented here s rules without very much justifiction. Other pproches cn be used. For exmple, common pproch is to begin with the Peno xioms the xioms of the nturl numbers nd build up to the rel numbers through severl completions of the nturl numbers. It s lso possible to begin with the xioms of set theory to build up the Peno xioms s theorems nd then use those to prove our xioms s further theorems. No mtter how it s done, there re lwys some xioms t the bse of the structure nd the rules for the rel numbers re the sme, whether they re xioms or theorems. We choose to strt t the top becuse the other pproches quickly turn into long nd tedious lbyrinth of technicl exercises without much connection to nlysis. 1. The Field Axioms These first six xioms re clled the field xioms becuse ny object stisfying them is clled field. They give the rithmetic properties of the rel numbers. A field is nonempty set F long with two binry opertions, multipliction : F F F nd ddition + : F F F stisfying the following xioms. 1 Axiom 1 (Associtive Lws). If, b, c F, then ( + b) + c = + (b + c) nd ( b) c = (b c). Axiom 2 (Commuttive Lws). If, b F, then + b = b + nd b = b. Axiom 3 (Distributive Lw). If, b, c F, then (b+c) = ( b)+( c). 1 Given set A, function f : A A A is clled binry opertion. In other words, binry opertion is just function with two rguments. The stndrd nottions of +(, b) = + b nd (, b) = b re used here. The symbol is unfortuntely used for both the Crtesin product nd the field opertion, but the context in which it s used removes the mbiguity. 2-1

25 2-2 CHAPTER 2. THE REAL NUMBERS Axiom 4 (Existence of identities). There re 0, 1 F with 0 1 such tht + 0 = nd 1 =, for ll F. Axiom 5 (Existence of n dditive inverse). For ech F there is F such tht + ( ) = 0. Axiom 6 (Existence of multiplictive inverse). For ech F \ {0} there is 1 F such tht 1 = 1. Although these xioms seem to contin most properties of the rel numbers we normlly use, they don t chrcterize the rel numbers; they just give the rules for rithmetic. There re mny other fields besides the rel numbers nd studying them is lrge prt of most bstrct lgebr courses. Exmple 2.1. From elementry lgebr we know tht the rtionl numbers Q = {p/q : p Z q N} form field. It is shown in Theorem 2.14 tht 2 / Q, so Q doesn t contin ll the rel numbers. Exmple 2.2. Let F = {0, 1, 2} with ddition nd multipliction clculted modulo 3. The ddition nd multipliction tbles re s follows It is esy to check tht the field xioms re stisfied. This field is usully clled Z 3. The following theorems, contining just few useful properties of fields, re presented mostly s exmples showing how the xioms re used. More complete developments cn be found in ny beginning bstrct lgebr text. Theorem 2.1. The dditive nd multiplictive identities of field F re unique. Proof. Suppose e 1 nd e 2 re both multiplictive identities in F. Then e 1 = e 1 e 2 = e 2, so the multiplictive identity is unique. The proof for the dditive identity is essentilly the sme. Theorem 2.2. Let F be field. If, b F with b 0, then nd b 1 re unique. Proof. Suppose b 1 nd b 2 re both multiplictive inverses for b 0. Then, using Axioms 4 nd 1, b 1 = b 1 1 = b 1 (b b 2 ) = (b 1 b) b 2 = 1 b 2 = b 2. This shows the multiplictive inverse in unique. The proof is essentilly the sme for the dditive inverse.

26 2. THE ORDER AXIOM 2-3 There re mny other properties of fields which could be proved here, but they correspond to the usul properties of the rel numbers lerned in elementry school, so we omit them. Some of them re in the exercises t the end of this chpter. From now on, the stndrd nottions for lgebr will usully be used; e. g., we will llow b insted of b, b for + ( b) nd /b insted of b 1. The reder my lso use the stndrd fcts she lerned from elementry lgebr. 2. The Order Axiom The xiom of this section gives the order nd metric properties of the rel numbers. In sense, the following xiom dds some geometry to field. Axiom 7 (Order xiom.). There is set P F such tht () If, b P, then + b, b P. 2 (b) If F, then exctly one of the following is true: P, P or = 0. Any field F stisfying the xioms so fr listed is nturlly clled n ordered field. Of course, the set P is known s the set of positive elements of F. Using Axiom 7(b), we see F is divided into three pirwise disjoint sets: P, {0} nd { x : x P }. The ltter of these is, of course, the set of negtive elements from F. The following definition introduces fmilir nottion for order. Definition 2.3. We write < b or b >, if b P. The menings of b nd b re now s expected. Notice tht > 0 = 0 P nd < 0 = 0 P, so > 0 nd < 0 gree with our usul notions of positive nd negtive. Our gol is to cpture ll the properties of the rel numbers with the xioms. The order xiom elimintes mny fields from considertion. For exmple, Exercise 2.7 shows the field Z 3 of Exmple 2.2 is not n ordered field. On the other hnd, fcts from elementry lgebr imply Q is n ordered field, so the first seven xioms still don t cpture the rel numbers. Following re few stndrd properties of ordered fields. Theorem 2.4. Let F be n ordered field nd F. 0 iff 2 > 0. Proof. ( ) If > 0, then 2 > 0 by Axiom 7(). If < 0, then > 0 by Axiom 7(b) nd 2 = 1 2 = ( 1)( 1) 2 = ( ) 2 > 0. ( ) Since 0 2 = 0, this is obvious. Theorem 2.5. If F is n ordered field nd, b, c F, then () < b + c < b + c, 2 Algebr texts would sy is P is closed under ddition nd multipliction. In Chpter 5 we ll use the word closed with different mening. This is one of the cses where lgebrists nd nlysts spek different lnguges. Fortuntely, the context usully erses confusion.

27 2-4 CHAPTER 2. THE REAL NUMBERS (b) < b b < c = < c, (c) < b c > 0 = c < bc, (d) < b c < 0 = c > bc. Proof. () < b b P (b + c) ( + c) P + c < b + c. (b) By supposition, both b, c b P. Using the fct tht P is closed under ddition, we see (b ) + (c b) = c P. Therefore, c >. (c) Since both b, c P nd P is closed under multipliction, c(b ) = cb c P nd, therefore, c < bc. (d) By ssumption, b, c P. Apply prt (c) nd Exercise 2.1. Theorem 2.6 (Two Out of Three Rule). Let F be n ordered field nd, b, c F. If b = c nd ny two of, b or c re positive, then so is the third. Proof. If > 0 nd b > 0, then Axiom 7() implies c > 0. Next, suppose > 0 nd c > 0. In order to force contrdiction, suppose b 0. In this cse, Axiom 7(b) shows which is impossible. Corollry > 0 Proof. Exercise ( b) = (b) = c < 0, Corollry 2.8. Let F be n ordered field nd F. If > 0, then 1 > 0. If < 0, then 1 < 0. Proof. The proof is Exercise 2.3. An ordered field begins to look like wht we expect for the rel numbers. The number line works pretty much s usul. Combining Corollry 2.7 nd Axiom 7(), it follows tht 2 = > 1 > 0, 3 = > 2 > 0 nd so forth. By induction, it is seen there is copy of N embedded in F. Similrly, there re lso copies of Z nd Q in F. This shows every ordered field is infinite. But, there might be holes in the line. For exmple if F = Q, numbers like 2, e nd π re missing. Definition 2.9. If F is n ordered field nd < b in F, then (, b) = {x F : < x < b}, (, ) = {x F : < x} nd (, ) = {x F : > x} re clled open intervls. (The ltter two re sometimes clled open right nd left rys, respectively.) The sets [, b] = {x F : x b} [, ) = {x F : x} nd (, ] = {x F : x} re clled closed intervls. (As bove, the ltter two re sometimes clled closed rys.)

28 2. THE ORDER AXIOM 2-5 [, b) = {x F : x < b} nd (, b] = {x F : < x b} re hlf-open intervls. The difference between the open nd closed intervls is tht open intervls don t contin their endpoints nd closed intervls contin their endpoints. In the cse of ry, the intervl only hs one endpoint. It is incorrect to write ry s (, ] or [, ] becuse neither nor is n element of F. The symbols nd re just plce holders telling us the intervls continue forever to the right or left Metric Properties. The order xiom on field F llows us to introduce the ide of the distnce between points in F. To do this, we begin with the following fmilir definition. Definition Let F be n ordered field. The bsolute vlue function on F is function : F F defined s { x, x 0 x = x, x < 0. The most importnt properties of the bsolute vlue function re contined in the following theorem. Theorem Let F be n ordered field nd x, y F. Then () x 0 nd x = 0 x = 0; (b) x = x ; (c) x x x ; (d) x y y x y; nd, (e) x + y x + y. Proof. () The fct tht x 0 for ll x F follows from Axiom 7(b). Since 0 = 0, the second prt is cler. (b) If x 0, then x 0 so tht x = ( x) = x = x. If x < 0, then x > 0 nd x = x = x. (c) If x 0, then x = x x = x. If x < 0, then x = ( x) = x < x = x. (d) This is left s Exercise 2.4. (e) Add the two sets of inequlities x x x nd y y y to see ( x + y ) x+y x + y. Now pply (d). This is usully clled the tringle inequlity. From studying nlytic geometry nd clculus, we re used to thinking of x y s the distnce between the numbers x nd y. This notion of distnce between two points of set cn be generlized. Definition Let S be set nd d : S S F stisfy () for ll x, y S, d(x, y) 0 nd d(x, y) = 0 x = y, (b) for ll x, y S, d(x, y) = d(y, x), nd

29 2-6 CHAPTER 2. THE REAL NUMBERS (c) for ll x, y, z S, d(x, z) d(x, y) + d(y, z). Then the function d is metric on S. The pir (S, d) is clled metric spce. A metric is function which defines the distnce between ny two points of set. Exmple 2.3. Let S be set nd define d : S S S by { 1, x y d(x, y) = 0, x = y. It cn redily be verified tht d is metric on S. This simplest of ll metrics is clled the discrete metric nd it cn be defined on ny set. It s not often useful. Theorem If F is n ordered field, then d(x, y) = x y is metric on F. Proof. Use prts (), (b) nd (e) of Theorem The metric on F derived from the bsolute vlue function is clled the stndrd metric on F. There re other metrics sometimes defined for specilized purposes, but we won t hve need of them. 3. The Completeness Axiom All the xioms given so fr re obvious from beginning lgebr, nd, on the surfce, it s not obvious they hven t cptured ll the properties of the rel numbers. Since Q stisfies them ll, the following theorem shows we re not yet done. Theorem There is no α Q such tht α 2 = 2. Proof. Assume to the contrry tht there is α Q with α 2 = 2. Then there re p, q N such tht α = p/q with p nd q reltively prime. Now, (2.1) ( ) p 2 = 2 = p 2 = 2q 2 q shows p 2 is even. Since the squre of n odd number is odd, p must be even; i. e., p = 2r for some r N. Substituting this into (2.1), shows 2r 2 = q 2. The sme rgument s bove estblishes q is lso even. This contrdicts the ssumption tht p nd q re reltively prime. Therefore, no such α exists. Since we suspect 2 is perfectly fine number, there s still something missing from the list of xioms. Completeness is the missing ide. The Completeness Axiom is somewht more complicted thn the previous xioms, nd severl definitions re needed in order to stte it.

30 3. THE COMPLETENESS AXIOM Bounded Sets. Definition A subset S of n ordered field F is bounded bove, if there exists M F such tht M x for ll x S. A subset S of n ordered field F is bounded below, if there exists m F such tht m x for ll x S. The elements M nd m re clled n upper bound nd lower bound for S, respectively. If S is bounded both bove nd below, it is bounded set. There is no requirement in the definition tht the upper nd lower bounds for set re elements of the set. They cn be elements of the set, but typiclly re not. For exmple, if S = (, 0), then [0, ) is the set of ll upper bounds for S, but none of them is in S. On the other hnd, if T = (, 0], then [0, ) is gin the set of ll upper bounds for T, but in this cse 0 is n upper bound which is lso n element of T. A set need not hve upper or lower bounds. For exmple S = (, 0) hs no lower bounds, while P = (0, ) hs no upper bounds. The integers, Z, hs neither upper nor lower bounds. If set hs no upper bound, it is unbounded bove nd, if it hs no lower bound, then it is unbounded below. In either cse, it is usully just sid to be unbounded. If M is n upper bound for the set S, then every x M is lso n upper bound for S. Considering some simple exmples should led you to suspect tht mong the upper bounds for set, there is one tht is best in the sense tht everything greter is n upper bound nd everything less is not n upper bound. This is the bsic ide of completeness. Definition Suppose F is n ordered field nd S is bounded bove in F. A number B F is clled lest upper bound of S if () B is n upper bound for S, nd (b) if α is ny upper bound for S, then B α. If S is bounded below in F, then number b F is clled gretest lower bound of S if () b is lower bound for S, nd (b) if α is ny lower bound for S, then b α. Theorem If F is n ordered field nd A F is nonempty, then A hs t most one lest upper bound nd t most one gretest lower bound. Proof. Suppose u 1 nd u 2 re both lest upper bounds for A. Since u 1 nd u 2 re both upper bounds for A, two pplictions of Definition 2.16 shows u 1 u 2 u 1 = u 1 = u 2. The proof of the other cse is similr. Definition If A F is nonempty nd bounded bove, then the lest upper bound of A is written lub A. When A is not bounded bove, we write lub A =. When A =, then lub A =.

31 2-8 CHAPTER 2. THE REAL NUMBERS If A F is nonempty nd bounded below, then the gretest lower bound of A is written glb A. When A is not bounded below, we write glb A =. When A =, then glb A =. 3 Notice the symbol is not n element of F. Writing lub A = is just convenient wy to sy A hs no upper bounds. Similrly lub = tells us hs every rel number s n upper bound. Theorem Let A F nd α F. α = lub A iff (α, ) A = nd for ll ε > 0, (α ε, α] A. Similrly, α = glb A iff (, α) A = nd for ll ε > 0, [α, α + ε) A. Proof. We will prove the first sttement, concerning the lest upper bound. The second sttement, concerning the gretest lower bound, follows similrly. ( ) If x (α, ) A, then α cnnot be n upper bound of A, which is contrdiction. If there is n ε > 0 such tht (α ε, α] A =, then from bove, we conclude = ((α ε, α] A) ((α, ) A) = (α ε, ) A. So, α ε/2 is n upper bound for A which is less thn α = lub A. This contrdiction shows (α ε, α] A. ( ) The ssumption tht (α, ) A = implies α lub A. On the other hnd, suppose lub A < α. By ssumption, there is n x (lub A, α] A. This is clerly contrdiction, since lub A < x A. Therefore, α = lub A. An egle-eyed reder my wonder why the intervls in Theorem 2.19 re (α ε, α] nd [α, α + ε) insted of (α ε, α) nd (α, α + ε). Just consider the cse A = {α} to see tht the theorem fils when the intervls re open. When lub A / A or glb A / A, the intervls cn be open, s shown in the following corollry. Corollry If A is bounded bove nd α = lub A / A, then for ll ε > 0, (α ε, α) A is n infinite set. Similrly, if A is bounded below nd β = glb A / A, then for ll ε > 0, (β, β + ε) A is n infinite set. Proof. Let ε > 0. According to Theorem 2.19, there is n x 1 (α ε, α] A. By ssumption, x 1 < α. We continue by induction. Suppose n N nd x n hs been chosen to stisfy x n (α ε, α) A. Using Theorem 2.19 s before to choose x n+1 (x n, α) A. The set {x n : n N} is infinite nd contined in (α ε, α) A. When F = Q, Theorem 2.14 shows there is no lest upper bound for A = {x : x 2 < 2} in Q. In sense, Q hs hole where this lest upper bound should be. Adding the following completeness xiom enlrges Q to fill in the holes. 3 Some people prefer the nottion sup A nd inf A insted of lub A nd glb A, respectively. They stnd for the supremum nd infimum of A.

32 3. THE COMPLETENESS AXIOM 2-9 Axiom 8 (Completeness). Every nonempty set which is bounded bove hs lest upper bound. This is the finl xiom. Any field F stisfying ll eight xioms is clled complete ordered field. We ssume the existence of complete ordered field, R, clled the rel numbers. In nive set theory it cn be shown tht if F 1 nd F 2 re both complete ordered fields, then they re the sme, in the following sense. There exists unique bijective function i : F 1 F 2 such tht i( + b) = i() + i(b), i(b) = i()i(b) nd < b i() < i(b). Such function i is clled n order isomorphism. The existence of such n order isomorphism shows tht R is essentilly unique. More reding on this topic cn be done in some dvnced texts [12, 13]. Every sttement bout upper bounds hs dul sttement bout lower bounds. A proof of the following dul to Axiom 8 is left s n exercise. Corollry Every nonempty subset of R which is bounded below hs gretest lower bound. In Section 4 it will be shown tht there is n x R stisfying x 2 = 2. This will show R removes the deficiency of Q highlighted by Theorem The Completeness Axiom plugs up the holes in Q Some Consequences of Completeness. The property of completeness is wht seprtes nlysis from geometry nd lgebr. Anlysis requires the use of pproximtion, infinity nd more dynmic visuliztions thn lgebr or clssicl geometry. The rest of this course is lrgely concerned with pplictions of completeness. Theorem 2.22 (Archimeden Principle ). If R, then there exists n N such tht n >. Proof. If the theorem is flse, then is n upper bound for N. Let β = lub N. According to Theorem 2.19 there is n m N such tht m > β 1. But, this is contrdiction becuse β = lub N < m + 1 N. Some other vritions on this theme re in the following corollries. Corollry Let, b R with > 0. () There is n n N such tht n > b. (b) There is n n N such tht 0 < 1/n <. (c) There is n n N such tht n 1 < n. Proof. () Use Theorem 2.22 to find n N where n > b/. (b) Let b = 1 in prt (). (c) Theorem 2.22 gurntees tht S = {n N : n > }. If n is the lest element of this set, then n 1 / S nd n 1 < n. Corollry If I is ny intervl from R, then I Q nd I Q c.

33 2-10 CHAPTER 2. THE REAL NUMBERS Proof. See Exercises 2.15 nd A subset of R which intersects every intervl is sid to be dense in R. Corollry 2.24 shows both the rtionl nd irrtionl numbers re dense. 4. Comprisons of Q nd R All of the bove still does not estblish tht Q is different from R. In Theorem 2.14, it ws shown tht the eqution x 2 = 2 hs no solution in Q. The following theorem shows x 2 = 2 does hve solutions in R. Since copy of Q is embedded in R, it follows, in sense, tht R is bigger thn Q. Theorem There is positive α R such tht α 2 = 2. Proof. Let S = {x > 0 : x 2 < 2}. Then 1 S, so S. If x 2, then Theorem 2.5(c) implies x 2 4 > 2, so S is bounded bove. Let α = lub S. It will be shown tht α 2 = 2. Suppose first tht α 2 < 2. This ssumption implies (2 α 2 )/(2α + 1) > 0. According to Corollry 2.23, there is n n N lrge enough so tht Therefore, 0 < 1 n < 2 α2 2α + 1 = 0 < 2α + 1 n < 2 α 2. ( α + 1 ) 2 = α 2 + 2α n n + 1 n 2 = α2 + 1 ( 2α + 1 ) n n < α 2 (2α + 1) + < α 2 + (2 α 2 ) = 2 n contrdicts the fct tht α = lub S. Therefore, α 2 2. Next, ssume α 2 > 2. In this cse, choose n N so tht 0 < 1 n < α2 2 2α = 0 < 2α n < α2 2. Then ( α 1 n ) 2 = α 2 2α n + 1 n 2 > α2 2α n > α2 (α 2 2) = 2, gin contrdicts tht α = lub S. Therefore, α 2 = 2. Theorem 2.14 leds to the obvious question of how much bigger R is thn Q. First, note tht since N Q, it is cler tht crd (Q) ℵ 0. On the other hnd, every q Q hs unique reduced frctionl representtion q = m(q)/n(q) with m(q) Z nd n(q) N. This gives n injective function f : Q Z N defined by f(q) = (m(q), n(q)), nd we conclude crd (Q) crd (Z N) = ℵ 0. The following theorem ensues. Theorem crd (Q) = ℵ 0.

34 4. COMPARISONS OF Q AND R 2-11 α 1 =.α 1 (1) α 1 (2) α 1 (3) α 1 (4) α 1 (5)... α 2 =.α 2 (1) α 2 (2) α 2 (3) α 2 (4) α 2 (5)... α 3 =.α 3 (1) α 3 (2) α 3 (3) α 3 (4) α 3 (5)... α 4 =.α 4 (1) α 4 (2) α 4 (3) α 4 (4) α 4 (5)... α 5 =.α 5 (1) α 5 (2) α 5 (3) α 5 (4) α 5 (5) Figure 2.1. The proof of Theorem 2.27 is clled the digonl rgument becuse it constructs new number z by working down the min digonl of the rry shown bove, mking sure z(n) α n (n) for ech n N. In 1874, Georg Cntor first showed tht R is not countble. The following proof is his fmous digonl rgument from Theorem crd (R) > ℵ 0 Proof. It suffices to prove tht crd ([0, 1]) > ℵ 0. If this is not true, then there is bijection α : N [0, 1]; i.e., (2.2) [0, 1] = {α n : n N}. Ech x [0, 1] cn be written in the deciml form x = n=1 x(n)/10n where x(n) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for ech n N. This deciml representtion is not necessrily unique. For exmple, 1 2 = 5 10 = n. In such cse, there is choice of x(n) so it is constntly 9 or constntly 0 from some N onwrd. When given choice, we will lwys opt to end the number with string of nines. With this convention, the deciml representtion of x is unique. Define z [0, 1] by choosing z(n) {d ω : d 8} such tht z(n) α n (n). Let z = n=1 z(n)/10n. Since z [0, 1], there is n n N such tht z = α n. But, this is impossible becuse z(n) differs from α n in the nth deciml plce. This contrdiction shows crd ([0, 1]) > ℵ 0. Around the turn of the twentieth century these then-new ides bout infinite sets were very controversil in mthemtics. This is becuse some of these ides re very unintuitive. For exmple, the rtionl numbers re countble set nd the irrtionl numbers re uncountble, yet between every two rtionl numbers re n uncountble number of irrtionl numbers nd between every two irrtionl numbers there re countbly infinite number of rtionl numbers. It would seem there re either too few or too mny gps in the sets to mke this possible. Such seemingly prdoxicl sitution flies in the fce of our intuition, which ws developed with finite sets in mind. n=2